The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $23$ years; the standard deviation is $5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $38$ years.
$23$ $18$ $28$ $13$ $33$ $8$ $38$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $23$ years. We know the standard deviation is $5$ years, so one standard deviation below the mean is $18$ years and one standard deviation above the mean is $28$ years. Two standard deviations below the mean is $13$ years and two standard deviations above the mean is $33$ years. Three standard deviations below the mean is $8$ years and three standard deviations above the mean is $38$ years. We are interested in the probability of a porcupine living less than $38$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the porcupines will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $8$ years and the other half $({0.15\%})$ will live longer than $38$ years. The probability of a particular porcupine living less than $38$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.